Question Number 191887 by Rupesh123 last updated on 03/May/23 Answered by mehdee42 last updated on 03/May/23 $${hop}\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \frac{{sinxcos}\mathrm{2}{xco}\mathrm{3}{x}…{cosnx}+\mathrm{2}{sin}\mathrm{2}{xcosxcos}\left(\mathrm{3}{x}\right)…{cos}\left({nx}\right)+…+{nsin}\left({nx}\right){cosxcos}\left(\mathrm{2}{x}\right)…{cos}\left({n}−\mathrm{1}\right){x}}{\mathrm{2}{x}}= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\left(\mathrm{1}+\mathrm{4}+\mathrm{9}+…+{n}^{\mathrm{2}} \right)}{{x}}=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\checkmark \\ $$ Commented…
Question Number 126349 by mnjuly1970 last updated on 19/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:{calculate}\:::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega\overset{???} {=}\int_{\mathrm{0}} ^{\:\:\infty} {e}^{\:−{t}} \:{t}^{\:\mathrm{2}} \:{j}_{\mathrm{0}} \left(\:{t}\:\right){dt} \\ $$$$\:\:\:\:\:{where}\::\:\:{j}_{\left({v}\right)} \left({x}\right)={x}^{{v}} \underset{{n}=\mathrm{0}} {\overset{\:\infty}…
Question Number 60812 by peter frank last updated on 26/May/19 Commented by tanmay last updated on 26/May/19 $${area}\:{of}\:{BCADB}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}×\mathrm{12}=\mathrm{60} \\ $$$$\mathrm{60}=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{13}×{h}\right] \\ $$$${h}=\frac{\mathrm{60}}{\mathrm{13}} \\ $$$${sin}\theta_{\mathrm{1}} =\frac{{h}}{\mathrm{12}}\:\:\:{and}\:{sin}\theta_{\mathrm{2}}…
Question Number 126344 by mnjuly1970 last updated on 19/Dec/20 $$\:\:\:\:\: \\ $$$$\:\:\:{evaluate}\:::\:\:\:\int_{−\mathrm{1}} ^{\:\:\mathrm{0}} \frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{x}^{\mathrm{3}} }}\:=? \\ $$$$ \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 60808 by naka3546 last updated on 26/May/19 $${Prove}\:\:{or}\:\:{disprove}\:\:{that}\:\:{there}\:\:{is} \\ $$$${a}\:\:{positive}\:\:{integer}\:\:{suitable}\:\:{for} \\ $$$$\:\:\:\:\:\:\:{n}^{\mathrm{3}} \:+\:\mathrm{1}\:\:\:\mid\:\:\:{n}!\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:{n}!\:\:\:{is}\:\:{divided}\:\:{by}\:\:{n}^{\mathrm{3}} \:+\:\mathrm{1}\:\:\right) \\ $$$${n}\:\:\in\:\:\mathbb{Z}^{+} \\ $$ Commented by mr W last…
Question Number 191873 by cherokeesay last updated on 02/May/23 Answered by mehdee42 last updated on 03/May/23 $${let}\:\::\:\angle{RPS}={p}_{\mathrm{1}} \:\&\angle\:{QPS}={p}_{\mathrm{2}} \\ $$$${p}_{\mathrm{1}} =\mathrm{45}−{x}\:\:\:,\:\:{p}_{\mathrm{2}} =\mathrm{135}−\mathrm{3}{x} \\ $$$$\frac{{PS}}{{QS}}=\frac{{sin}\mathrm{3}{x}}{{sinp}_{\mathrm{2}} }=\frac{{sinx}}{{sinp}_{\mathrm{1}}…
Question Number 191874 by Subhi last updated on 02/May/23 $${lim}_{{x}\Rightarrow\infty} \frac{{e}^{{x}+\mathrm{1}} +{pi}^{{x}−\mathrm{1}} }{{e}^{{x}−\mathrm{1}} +{pi}^{{x}+\mathrm{1}} } \\ $$$$ \\ $$ Answered by mehdee42 last updated on…
Question Number 191868 by Spillover last updated on 02/May/23 $${A}\:{particle}\:{of}\:{mass}\:{m}\:{moves}\:{under}\:{the}\:{central} \\ $$$${repulsive}\:{force}\:\frac{{mb}}{{r}^{\mathrm{3}} }\:\:{and}\:{is}\:{initially}\:{moving} \\ $$$${at}\:{a}\:{distance}\:'{a}'\:\:{from}\:{the}\:{origin}\:{of}\:\:{a}\:{force} \\ $$$${with}\:{velocity}\:\:'{v}'\:{at}\:{right}\:{angle}\:{to}\:\:'{a}'. \\ $$$${show}\:{that}\:\:\: \\ $$$$\:\:\:\:\:{r}\mathrm{cos}\:{p}\theta={a}\:\:{where}\:{p}\:=\frac{{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} }+\mathrm{1}. \\ $$$$…
Question Number 60797 by arcana last updated on 25/May/19 $$\int\frac{{e}^{{w}} }{{w}^{{n}+\mathrm{1}} }{dw},\:{n}\in\mathbb{N} \\ $$ Commented by MJS last updated on 26/May/19 $$\mathrm{this}\:\mathrm{reminds}\:\mathrm{me}\:\mathrm{of}\:\Gamma\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{t}} {t}^{{x}−\mathrm{1}}…
Question Number 191867 by Spillover last updated on 02/May/23 $${Prove}\:{that}\:{if}\:\:\:{u}={f}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right),{where}\:{f}\:\:{is}\:{arbitry} \\ $$$${function}\:{then}\:\:\:\:{x}^{\mathrm{2}} \:\frac{\partial{u}}{\partial{y}}\:=\:{y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$ Answered by qaz last updated on 02/May/23…